Question:
Let $p(x)$ be a polynomial with positive leading coefficient such that
\[[p(x)]^2 = 4(x^2 + 2x + 1)(x^2 + 3x - 2) + (x - 3)^2.\]Find $p(x).$

Answer:
Expanding, we get
\[[p(x)]^2 = 4x^4 + 20x^3 + 21x^2 - 10x + 1.\]Then $p(x)$ is quadratic, with leading term $2x^2.$  Let
\[p(x) = 2x^2 + bx + c.\]Then
\[[p(x)]^2 = 4x^4 + 4bx^3 + (b^2 + 4c) x^2 + 2bcx + c^2.\]Matching coefficients, we get
\begin{align*}
4b &= 20, \\
b^2 + 4c &= 21, \\
2bc &= -10, \\
c^2 &= 1.
\end{align*}From $4b = 20,$ $b = 5.$  Then from $2bc = -10,$ $c = -1.$  Hence, $p(x) = \boxed{2x^2 + 5x - 1}.$